372 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th7-2.pm5

Example 7.19. An ice skate is able to glide over the ice because the skate blade exerts

sufficient pressure on the ice that a thin layer of ice is melted. The skate blade then glides over

this thin melted water layer. Determine the pressure an ice skate blade must exert to allow

smooth ice skate at – 10°C.

The following data is given for the range of temperatures and pressures involved :

hfg(ice) = 334 kJ/kg ; vliq. = 1 × 10 m^3 /kg ; vice = 1.01 × 103 m^3 /kg.

Solution. Since it is a problem of phase change from solid to liquid, therefore, we can use

Clausius-Claperyon equation given below :

dp

dT

=

h

vT

fg

fg

.^1
Multiplying both the sides by dT and integrating, we get
dp
p

p

1

2

z =

h

v

dT

T

fg

fg T

T

1

2

z

or (p 2 – p 1 ) =

h

v

fg

fg

loge TT^2

1

F

HG

I

KJ ...(i)

But at p 1 = 1 atm., t 1 = 0°C

Thus, p 1 = 1.013 bar, T 1 = 0 + 273 = 273 K

p 2 = ?, T 2 = – 10 + 273 = 263 K

Substituting these values in eqn. (i), we get

(p 2 – 1.013 × 10^5 ) =

334 10

101

×^3

bg−1. × loge^

263

273

F

H

I

K

= 334 10 001

×^3

. × loge^

273

263

F

H

I

K = 12.46 × 10

(^5) N/m 2
or p 2 = 12.46 × 10^5 + 1.013 × 10^5
= 13.47 × 10^5 N/m^2 or 13.47 bar. (Ans.)
This pressure is considerably high. It can be achieved with ice skate blade by having only a
small portion of the blade surface in contact with the ice at any given time. If the temperature
drops lower than – 10°C, say – 15°C, then it is not possible to generate sufficient pressure to melt
the ice and conventional ice skating will not be possible.
Example 7.20. For mercury, the following relation exists between saturation pressure
(bar) and saturation temperature (K) :
log 10 p = 7.0323 – 3276.6/T– 0.652 log 10 T
Calculate the specific volume vg of saturation mercury vapour at 0.1 bar.
Given that the latent heat of vapourisation at 0.1 bar is 294.54 kJ/kg.
Neglect the specific volume of saturated mercury liquid.
Solution. Latent heat of vapourisation, hfg = 294.54 kJ/kg (at 0.1 bar) ...(given)
Using Clausius-Claperyon equation
dp
dT

h
vT
fg
fg

h
vvT
fg
()g− f
...(i)
Since vf is neglected, therefore eqn. (i) becomes
dp
dT

h
vT
fg
g
Now, log 10 p = 7.0323 – 3276 6T. – 0.652 log 10 T